Answer
$y=2cos(t)-sin(t)$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''-{y}''+{y}'-y=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}-r^2e^{rt}+re^{rt}-e^{rt}=0\\\\$
$r^3-r^2+r+-1=(r-1)(r^2+1)=0 \rightarrow\;\;\;\;\; r_{1}= 1\;\;\;\;\;or\;\;\;r_{2}=i\;,r_{3}=-i \;\;\;\;\;\;\\\\$
So the 3 roots are: $\;\;\;\;\;\;r_{1}=1 \;\;\;,\;\;r_{2},r_{3}=\pm i $
The general solution for complex roots is:
$y= C_{1}e^{\alpha t}cos(\beta t)+C_{2}e^{\alpha t}sin(\beta t)$
$y= C_{1}e^{t}+C_{2}cos(t)+C_{3}sin(t)$
Derivatives of the general solution;
${y}'=C_{1}e^t-C_{2}sin(t)+C_{3}cos(t)$
${y}''=C_{1}e^t-C_{2}cos(t)-C_{3}sin(t)$
At;
$y(0)=C_{1}+C_{2}=2 \;\;\;\;\;\Rightarrow \;\;\;\;\;C_{2}=2-C_{1} $
${y}'(0)=C_{1}+C_{3}=-1 \;\;\;\;\;\Rightarrow \;\;\;\;\;C_{3}=-1-C_{1} $
${y}''(0)=C_{1}-C_{2}=-2 $
$\;\;\;\;\;\Rightarrow \;\;\;\;C_{1}=0\;\;\;\;\;\;\;C_{3}=-1\;\;\;\;\;\;C_{2}=2$
$\therefore y=2cos(t)-sin(t)$
